POJ--3723 Conscription(最小生成树)

记录
18:30 2023-1-29

http://poj.org/problem?id=3723

reference:《挑战程序设计竞赛(第2版)》2.5.6 p109

Description

Windy has a country, and he wants to build an army to protect his country. He has picked up N girls and M boys and wants to collect them to be his soldiers. To collect a soldier without any privilege, he must pay 10000 RMB. There are some relationships between girls and boys and Windy can use these relationships to reduce his cost. If girl x and boy y have a relationship d and one of them has been collected, Windy can collect the other one with 10000-d RMB. Now given all the relationships between girls and boys, your assignment is to find the least amount of money Windy has to pay. Notice that only one relationship can be used when collecting one soldier.

Input

The first line of input is the number of test case.
The first line of each test case contains three integers, N, M and R.
Then R lines followed, each contains three integers xi, yi and di.
There is a blank line before each test case.

1 ≤ N, M ≤ 10000
0 ≤ R ≤ 50,000
0 ≤ xi < N
0 ≤ yi < M
0 < di < 10000

Output

For each test case output the answer in a single line.

Sample Input

2

5 5 8
4 3 6831
1 3 4583
0 0 6592
0 1 3063
3 3 4975
1 3 2049
4 2 2104
2 2 781

5 5 10
2 4 9820
3 2 6236
3 1 8864
2 4 8326
2 0 5156
2 0 1463
4 1 2439
0 4 4373
3 4 8889

Sample Output

71071
54223

最小生成树问题。要想让花费的钱最少，也就是尽可能选择关系最大的边，这样转换成了一个最大生成树问题，将边上的值取成负号就变成了最小生成树问题。(怕什么真理无穷，进一寸有一寸的惊喜)

#include<cstdio>
#include<vector>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
#define MAX_V 100000
#define MAX_E 100000
#define MAX_R 100000

const int INF = 0x3f3f3f3f;
#define MAX_N 100000
int par[MAX_N];
int ran[MAX_N];

void init(int n) {
    for(int i = 0; i < n; i++) {
        par[i] = i;
        ran[i] = 0;
    }
}

int find(int x) {
    if(par[x] == x) {
        return x;
    } else {
        return par[x] = find(par[x]);
    }
}

void unite(int x, int y) {
    x = find(x);
    y = find(y);

    if(x == y) return;

    if(ran[x] < ran[y]) {
        par[x] = y;
    } else {
        par[y] = x;
        if(ran[x] == ran[y]) ran[x]++;
    }
}

bool same(int x, int y) {
    return find(x) == find(y);
}

struct edge {int u, v, cost;};

bool comp(const edge &e1, const edge &e2) {
    return e1.cost < e2.cost;
}

edge es[MAX_E];
int V, E;

int kruskal() {
    sort(es, es + E, comp); // 按照edge.cost的顺序 从小到大
    init(V);
    int res = 0;
    for(int i = 0; i < E; i++) {
        edge e = es[i];
        if(!same(e.u, e.v)) {
            unite(e.u, e.v);
            res += e.cost;
        }
    }
    return res;
}

int N, M, R;
int x[MAX_R], y[MAX_R], d[MAX_R];

void solve() {
    V = N + M;
    E = R;
    for(int i = 0; i < R; i++) {
        // es[i] = (edge){x[i], N + y[i], -d[i]};
        es[i].u = x[i];
        es[i].v = N + y[i];
        es[i].cost = -d[i];
    }
    printf("%d\n", 10000 * (N + M) + kruskal());
}

int main() {
    int T;
    scanf("%d", &T);
    for(int i = 0; i < T; i++) {
        scanf("%d %d %d", &N, &M, &R);
        for(int j = 0; j < R; j++) {
            scanf("%d %d %d", &x[j], &y[j], &d[j]);
        }
        solve();
    }
}